# Prove that the polar lines of three points outside a circle $C$, are concurrent if and only if the three points are colinear.

I want to know if the first part of my prove (the "if" part) is right. If not, could you show my mistakes please? Let the circle $$C$$ be given by the equation $$C: (\mathbf{x}-\mathbf{c})\cdot (\mathbf{x}-\mathbf{c})=r^2,$$ where $$\mathbf{c}$$ is the center of the circle and $$r$$ the radius. A line in its parametric form can be given by $$L= \{\mathbf{u}t + \mathbf{d} |t \in\mathbb{R}\}$$ Therefore, if we choose two points in the line $$L$$ with values $$t$$ equal to $$s$$ and $$f$$, the polar lines of the points are
$$S: (\mathbf{u}s + \mathbf{d}- \mathbf{c}) \cdot (\mathbf{x} - \mathbf{c}) = r^2,$$ $$F: (\mathbf{u}f + \mathbf{d}- \mathbf{c}) \cdot (\mathbf{x} - \mathbf{c}) = r^2 .$$ By substracting the two equations, we obtain $$(\mathbf{x} - \mathbf{c})\cdot((\mathbf{u}s + \mathbf{d}- \mathbf{c}) - (\mathbf{u}f + \mathbf{d}- \mathbf{c})) = 0$$ $$(\mathbf{x} - \mathbf{c})\cdot (s-f)\mathbf{u} = (\mathbf{x} - \mathbf{c}) \cdot \mathbf{u} = 0,$$ which is the equation of the line normal to $$L$$, and passing through the center of the circle $$\mathbf{c}$$, let this line be $$R$$. Since the values of $$s$$ and $$f$$ are arbitrary, let $$s$$ fixed and vary $$f$$, therefore the polar lines of the points of the form $$\mathbf{u}f + \mathbf{d}- \mathbf{c}$$ must past through the intersection of $$S$$ and $$R$$, in other words, they are concurrent.

The polar line of the point C is the red one

• You were doing find until the end. Although the intersection of the two polars that you’ve constructed does lie on $R$, it’s not the intersection of $L$ and $R$. Indeed, this point lies in the interior of the circle.
– amd
May 2, 2018 at 6:22
• Is it correct now? May 2, 2018 at 14:43

Let $\mathbf p$ and $\mathbf q$ be two distinct points exterior to $C$. Every point $\mathbf r$ colinear with them can be written in the form $(1-\lambda)\mathbf p+\lambda\mathbf q$ for some $\lambda\in\mathbb R$. An equation of the polar line of $\mathbf r$ is $$\left((1-\lambda)\mathbf p+\lambda\mathbf q-\mathbf c\right)\cdot(\mathbf x-\mathbf c)-r^2 = 0.$$ The left-hand side can be rewritten as $$(1-\lambda)[(\mathbf p-\mathbf c)\cdot(\mathbf x-\mathbf c)-r^2]+\lambda[(\mathbf q-\mathbf c)\cdot(\mathbf x-\mathbf c)-r^2],$$ so it is a linear combination of equations for the polars of $\mathbf p$ and $\mathbf q$, therefore passes through their intersection.
If you’re familiar with projective geometry and determinants, this property of pole/polar relationships is pretty much a direct consequence of the definitions. Both colinearity and concurrence can be defined by $\det\small{\begin{bmatrix}a&b&c\end{bmatrix}}=0$. If $C$ is the matrix of a non-degenerate conic, the polar of a point $\mathbf p$ is $C\mathbf p$ and the pole of a line $\mathbf l$ is $C^{-1}\mathbf l$. However, for any matrix $M$, $\det\small{\begin{bmatrix}Ma&Mb&Mc\end{bmatrix}} = \det(M) \det\small{\begin{bmatrix}a&b&c\end{bmatrix}}$, so three points are colinear iff their polar lines are concurrent.